ACM程序设计选修课——1040: Alex and Asd fight for two pieces of cake（YY+GCD）

1040: Alex and Asd fight for two pieces of cake

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 27  Solved: 12
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Description

Alex and Asd have found two pieces of cake on table of weight a and b grams.They are so greedy that they all want the larger piece. A fight may happenes.

Now the smart person-Radical comes in and starts the dialog: "Stupid people, wait a little, I will make your pieces equal" 

"Wow,you are so amazing, how are you going to do that?", Alex and Asd ask. Radical says"Ok,listen to me,If the mass of a certain piece is divisible by two, then I can eat exactly a half of the piece. 

If the mass of a certain piece is divisible by three, then I can eat exactly two-thirds, and if the mass is divisible by five, 

then I can eat four-fifths. I'll eat a little here and there and make the pieces equal".

Alougth they are not som smart , they got it.So they agrees to his proposal, but on one condition: Radical should make the pieces equal as quickly as possible. 

Find the minimum number of operations Radical needs to make pieces equal.

 

 

Input

The first line contains two space-separated integers a and b (1 ≤ a, b ≤ 109).

 

 

 

Output

If it is impossible to make the pieces equal, print -1. Otherwise, print the required minimum number of operations. If the pieces of the cake are initially equal, the required number is 0.

 

Sample Input

36 30
7 8
11 11

Sample Output

3
-1
0

题意：两个人的一定要分到相等的蛋糕，否则输出-1，若初始值就相等， 输出0。跟狐狸给两只熊分饼一个道理，每次吃掉1/2或2/3或4/5。

那么此题就可以理解为每次将初始值乘以1/2或1/3或1/5，Alex和Asd乘以这几个数的次数可以不一样，每次乘的值也可以不一样，求最少的次数让这两个人相等。

首先感觉是贪心，但是后来感觉2、3、5都是质数，2^a和3^b和5^c次的公因数都是1，应该不是贪心。

比如例一、36与30，gcd为6，6/36=1/6，6/30=1/5。

再进一步，题意就成了用1/2，1/3，1/5来凑gcd(Alex,Asd)/Alex（或Asd，可行状态下这两个假分数肯定相等且最简式分子为1）且项数最少。

再进一步，就是求上述分母分解为2、3、5的个数（感觉由于三个数互质，只有唯一解，不存在最大最小的问题。）

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
LL list[3]={5,3,2};//为了循环方便用个数组
LL gcd(LL a,LL b)
{
	return b?gcd(b,a%b):a;
}
int main (void)
{
	LL a,b;
	while (cin>>a>>b)
	{
		LL g=gcd(a,b);
		LL ca,cb,ag,bg,fenzia,fenzib,fenmua,fenmub;
		if(a==b)
		{
			cout<<0<<endl;
			continue;
		}
		else
		{
			map<LL,LL>lista;//记录Alex分母的分解情况
			map<LL,LL>listb;//记录Asd分母的分解情况
			ag=gcd(a,g);
			bg=gcd(b,g);
			fenmua=a/ag;//得到Alex最简分式的分母
			fenmub=b/bg;//得到Asd最简分式的分母
			for (int i=0; i<3; i++)//Alex分解
			{
				while (fenmua>=list[i])
				{
					if(fenmua%list[i]==0)
					{
						fenmua/=list[i];
						lista[list[i]]++;
					}
					else
						break;
				}
			}
			for (int i=0; i<3; i++)//Asd分解
			{
				while (fenmub>=list[i])
				{
					if(fenmub%list[i]==0)
					{
						fenmub/=list[i];
						listb[list[i]]++;
					}
					else
						break;
				}
			}
			if(fenmua==1&&fenmub==1)
				cout<<lista[2]+lista[3]+lista[5]+listb[2]+listb[3]+listb[5]<<endl;//输出操作次数(Alex+Asd)
			else
				cout<<-1<<endl;			
		}
	}
	return 0;
}